In thea diagram,ABCD is a trapezium with AB,DC and vertices A(-4;1),B(x;-3),C(1;y) and D.Determine the value of x if the distance AB is 5 units.
A(-4,1), B(x,-3).
(AB)^2 = (x-(-4))^2 + (-3-1)^2 = 5^2,
(x+4)^2 + 16 = 25 - 16 = 9 ,
Take sqrt of both sides:
x+4 = +- 3,
x = -4 +-3,
x = -1, and x = -7.